in triangle pqr, if qs is the angle bisector of angle q then show that A(triangle pqr)/A(triangle pqs)=pq/qr
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in Δ pqr if qs bisect ∠q then Area of Triangle sqr / Area of Triangle pqs = qr/pq
Step-by-step explanation:
Correct Question : A(triangle sqr)/A(triangle pqs)= qr/pq
Let say Angle q = 2β
Then qs is bisector of ∠q
=> ∠pqs = ∠rqs = β
now Draw sm ⊥ pq & sn ⊥ qr
Area of Triangle pqs
= (1/2) * pq * sm
in Δ smq
sinβ = sm/qs
=> sm = qs * sinβ
Area of Triangle pqs = (1/2) * pq *qs * sinβ
Area of Triangle sqr
= (1/2) * qr * sn
in Δ snq
sinβ = sn/qs
=> sn = qs * sinβ
Area of Triangle sqr = (1/2) * qr *qs * sinβ
Area of Triangle sqr / Area of Triangle pqs = (1/2) * qr *qs * sinβ / (1/2) * pq *qs * sinβ
=> Area of Triangle sqr / Area of Triangle pqs = qr/pq
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